blob: 3bb0fda472afeeeed277b516d5ea38a266147078 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
|
// Exa 3.1
clc;
clear;
// Given data
//Fig. 3.1(e) shows an inverting amplifier
ACL = -10; // open loop gain of ap-amp 741
R1 = 10*10^6; // Input impedence in Ω
// Solution
printf(' In fig. 3.1(e), to set input impedence Ri = 10 MΩ , pick R1 = 10 MΩ. ');
// since, ACL = - Rf/R1;
// Therefore,
Rf = -ACL*R1;
printf('\n The calculated value of Rf = %d MΩ. \n ',Rf/10^6);
printf(' Choose Rt = 47 kΩ. \n ');
Rt = 47*10^3; // Ω
Rs = (Rt^2)/(Rf-2*Rt);
printf(' Calculated Rs = %d Ω. ',Rs);
|